# Consider the expression. I = \int^1_0 \int^{y(2 - y)}_0 \int^{1 - y}_0 dx dz dy + \int ^1_0...

## Question:

Consider the expression.

{eq}I = \int^1_0 \int^{y(2 - y)}_0 \int^{1 - y}_0 dx dz dy + \int ^1_0 \int^1_{y(2 - y)} \int^{\sqrt {1 - y}}_0 dx dz dy {/eq}

Rewrite I using the order of integration dz dx dy.

## Double Integrals:

Recall that whenever we set up a triple integral, there are actually *six* different ways to do it, one for each order of integration. Choosing an order that produces the simplest integral is usually the name of the game, and there are many situations where changing from one order to another is just as simple as that.

## Answer and Explanation:

We have

{eq}\begin{align*} \int^1_0 \int^{y(2 - y)}_0 \int^{1 - y}_0 dx\ dz\ dy + \int ^1_0 \int^1_{y(2 - y)} \int^{\sqrt {1 - y}}_0 dx \ dz\ dy \end{align*} {/eq}

We are in luck because both {eq}x {/eq} and {eq}z {/eq} depend only on {eq}y {/eq}. Since this is the case, and the integrand is simply 1 and so doesn't depend on {eq}x {/eq} or {eq}z {/eq}, we can switch the the order without changing anything at all:

{eq}\begin{align*} \int^1_0 \int^{1 - y}_0 \int^{y(2 - y)}_0 dz\ dx\ dy + \int ^1_0 \int^{\sqrt {1 - y}}_0 \int^1_{y(2 - y)} dz \ dx\ dy \end{align*} {/eq}

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from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14